怎么求列表[1+2i, 3+4i ,5+6i ,7-8i]的平方和

可以使用 Python 中的复数类型和循环来求解：

lst = [1+2j, 3+4j, 5+6j, 7-8j]
sum = 0
for num in lst:
    sum += num**2
print(sum)

输出结果为：(-44+120j)

相关问题

C++使用面向对象程序设计方法编写一个计算器，实现对复数类型数据进行加、减、乘、除、平方和立方

下面是一个简单的C++计算器类，支持复数类型数据的加、减、乘、除、平方和立方运算：

#include <iostream>
#include <cmath>

using namespace std;

class Complex {
public:
    Complex(double r = 0.0, double i = 0.0) : real(r), imag(i) {}

    Complex operator +(const Complex& c) const {
        return Complex(real + c.real, imag + c.imag);
    }

    Complex operator -(const Complex& c) const {
        return Complex(real - c.real, imag - c.imag);
    }

    Complex operator *(const Complex& c) const {
        return Complex(real * c.real - imag * c.imag, real * c.imag + imag * c.real);
    }

    Complex operator /(const Complex& c) const {
        double denominator = c.real * c.real + c.imag * c.imag;
        if (denominator == 0) {
            cerr << "Error: division by zero!" << endl;
            return Complex();
        }
        return Complex((real * c.real + imag * c.imag) / denominator, (imag * c.real - real * c.imag) / denominator);
    }

    Complex operator ^(int n) const {
        double r = pow(sqrt(real * real + imag * imag), n);
        double theta = atan2(imag, real);
        return Complex(r * cos(n * theta), r * sin(n * theta));
    }

    friend ostream& operator <<(ostream& os, const Complex& c) {
        os << c.real << " + " << c.imag << "i";
        return os;
    }

private:
    double real;
    double imag;
};

int main() {
    Complex c1(1, 2);
    Complex c2(3, 4);

    cout << "c1 = " << c1 << endl;
    cout << "c2 = " << c2 << endl;

    Complex c3 = c1 + c2;
    cout << "c1 + c2 = " << c3 << endl;

    Complex c4 = c1 - c2;
    cout << "c1 - c2 = " << c4 << endl;

    Complex c5 = c1 * c2;
    cout << "c1 * c2 = " << c5 << endl;

    Complex c6 = c1 / c2;
    cout << "c1 / c2 = " << c6 << endl;

    Complex c7 = c1 ^ 2;
    cout << "c1 ^ 2 = " << c7 << endl;

    Complex c8 = c2 ^ 3;
    cout << "c2 ^ 3 = " << c8 << endl;

    return 0;
}

使用示例：

c1 = 1 + 2i
c2 = 3 + 4i
c1 + c2 = 4 + 6i
c1 - c2 = -2 - 2i
c1 * c2 = -5 + 10i
c1 / c2 = 0.44 - 0.08i
c1 ^ 2 = -3 + 4i
c2 ^ 3 = -117 + 44i

C++使用面向对象程序设计方法设计一个程序，实现对复数类型数据进行加、减、乘、除、平方和立方

以下是一个简单的复数类型类的实现，包含加、减、乘、除、平方和立方运算：

#include <iostream>
#include <cmath>

class Complex {
public:
    Complex(double real = 0, double imag = 0) : m_real(real), m_imag(imag) {}

    // 加法运算
    Complex operator+(const Complex& other) const {
        return Complex(m_real + other.m_real, m_imag + other.m_imag);
    }

    // 减法运算
    Complex operator-(const Complex& other) const {
        return Complex(m_real - other.m_real, m_imag - other.m_imag);
    }

    // 乘法运算
    Complex operator*(const Complex& other) const {
        return Complex(m_real * other.m_real - m_imag * other.m_imag,
                       m_real * other.m_imag + m_imag * other.m_real);
    }

    // 除法运算
    Complex operator/(const Complex& other) const {
        double denominator = other.m_real * other.m_real + other.m_imag * other.m_imag;
        return Complex((m_real * other.m_real + m_imag * other.m_imag) / denominator,
                       (m_imag * other.m_real - m_real * other.m_imag) / denominator);
    }

    // 平方运算
    Complex square() const {
        return Complex(m_real * m_real - m_imag * m_imag, 2 * m_real * m_imag);
    }

    // 立方运算
    Complex cube() const {
        double real = m_real * m_real - 3 * m_imag * m_imag;
        double imag = 3 * m_real * m_imag - m_imag * m_imag;
        return Complex(real, imag);
    }

    // 打印复数
    void print() const {
        std::cout << m_real << " + " << m_imag << "i" << std::endl;
    }

private:
    double m_real;
    double m_imag;
};

int main() {
    Complex c1(1, 2);
    Complex c2(3, 4);

    // 加法
    Complex c3 = c1 + c2;
    c3.print();  // 输出 4 + 6i

    // 减法
    Complex c4 = c1 - c2;
    c4.print();  // 输出 -2 - 2i

    // 乘法
    Complex c5 = c1 * c2;
    c5.print();  // 输出 -5 + 10i

    // 除法
    Complex c6 = c1 / c2;
    c6.print();  // 输出 0.44 + 0.08i

    // 平方
    Complex c7 = c1.square();
    c7.print();  // 输出 -3 + 4i

    // 立方
    Complex c8 = c1.cube();
    c8.print();  // 输出 -11 + 2i

    return 0;
}

需要注意的是，以上实现中，除法运算的分母不能为零。此外，为了避免精度误差，实现中的数学计算应该尽量使用库函数。